math185f09-hw4sol - MATH 185: COMPLEX ANALYSIS FALL 2009/10...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 4 SOLUTIONS 1. Let Ω ⊆ C be a region. Let f = u + iv be analytic on Ω. (a) If αu + βv is constant on Ω for some α,β ∈ C × , show that f is constant on Ω. Solution. Taking partial derivatives of αu + βv = constant , we get ( αu x + βv x = 0 , αu y + βv y = 0 . Substituting the Cauchy-Riemann equations u x = v y and u y =- v x gives ( αu x- βv y = 0 , αu y + βu x = 0 . Taking complex conjugate of the second equation (and noting that u x and u y are real valued) and writing the resulting system in matrix form yields α- β β α u x u y = . Since α,β 6 = 0, det α- β β α = | α | 2 + | β | 2 6 = 0 , and thus the system has a unique solution u x = 0, u y = 0. Applying the Cauchy-Riemann equations again, we get v x = 0, v y = 0. Since Ω is a region, we must have that f is constant-valued on Ω. (b) If u 2 + v 2 is constant on Ω, show that f is constant on Ω. Solution. Let the constant be c . Then u 2 + v 2 = c. If c = 0, then u = 0 and v = 0 and so f = 0, thus constant. We will assume that c 6 = 0. Taking partial derivatives, we get ( uu x + vv x = 0 , uu y + vv y = 0 . (1.1) Substituting the Cauchy-Riemann equations u x = v y and u y =- v x into the second equation of (5.12) gives ( uu x + vv x = 0 ,- uv x + vu x = 0 . (1.2) Now we multiply the first equation in (1.2) by u on both sides and use the second equation in (1.2) to get 0 = u ( uu x + vv x ) = u 2 u x + v ( uv x ) = u 2 u x + v ( vu x ) = ( u 2 + v 2 ) u x = cu x . Date : October 21, 2009 (Version 1.0). 1 Similarly, we could have instead substitute the Cauchy-Riemann equations u x = v y and u y =- v x into the first equation of (5.12) to get ( uv y- vu y = 0 , uu y + vv y = 0 . (1.3) This time, multiplying the second equation in (4.10) by v on both sides and use the first equation in (4.10) yields 0 = v ( uu y + vv y ) = u ( vu y ) + v 2 v y = u ( uv y ) + v 2 v y = ( u 2 + v 2 ) v y = cv y . Since c 6 = 0, we get u x = 0 and v y = 0 . Applying the Cauchy-Riemann equations again, we get v x = 0, u y = 0. Since Ω is a region, we must have that f is constant-valued on Ω. (c) If u = v 2 , show that f is constant on Ω. Solution. Applying the Cauchy-Riemann equations and chain rule, we get 2 vv x = v y , 2 vv y =- v x . Eliminating v x from the equations, we get- 4 v 2 v y = v y . Suppose v y 6≡ 0. Then there must exist a point z = x + iy ∈ Ω such that v y ( x ,y ) 6 = 0. So v ( x ,y ) 2 =- 1 / 4 — which is impossible since v is real-valued. Hence we must have v y ≡ 0 and in which case, we apply the Cauchy-Rieman equations again to get v x ≡ v y ≡ u x ≡ u y ≡ . Hence f is constant. (d) If u = ϕ ◦ v for some differentiable real function ϕ : R → R , show that f is constant on Ω....
View Full Document

This note was uploaded on 02/17/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.

Page1 / 9

math185f09-hw4sol - MATH 185: COMPLEX ANALYSIS FALL 2009/10...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online